Question: $g(x)=\dfrac{9}{\sqrt[{3}]{x+1}}$. On which intervals is the graph of $g$ concave down? Choose 1 answer: Choose 1 answer: (Choice A) A $(-1,1)$ only (Choice B) B $(1,\infty)$ only (Choice C) C $(-\infty,-1)$ only (Choice D) D $(0,\infty)$ only
Explanation: We can analyze the intervals where $g$ is concave up/down by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=\dfrac{4}{\sqrt[3]{(x+1)^7}}$. $g''$ is never equal to $0$. $g''$ is undefined for $x=-1$. Therefore, our only point of interest is $x=-1$. Our point of interest divides the domain of $g$ (which is all numbers except for $-1$ ) into two intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $(-\infty, \llap{-}1)$ $( \llap{-}1,\infty)$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $(-\infty,-1)$ $x=-2$ $g''(-2)=-4<0$ $g$ is concave down $\cap$ $(-1,\infty)$ $x=0$ $g''(0)=4>0$ $g$ is concave up $\cup$ In conclusion, the graph of $g$ is concave down over the interval $(-\infty,-1)$ only.